A subset $U\in\mathbb{R}^2$ is called radially open if at every point $\textbf{x}\in U$, $U$ contains an open segment through $\textbf{x}$ in every direction. That is, for every unit vector $\textbf{u} = \langle\cos(\theta), \sin(\theta)\rangle$, there exists $\epsilon>0$ such that $\textbf{x}+s\textbf{u}\in U$ for every $s\in (-\epsilon,\epsilon)$. Prove that the family of radially open sets $\tau$ in $\mathbb{R}^2$ is a topology.
Here's the proof I wrote. I am most unsure about the arbitrary union part of my proof. Any assistance would be appreciated, even if it's just a "Yep, that looks right". If my proof is wrong, I would prefer hints as opposed to full answers.
- $\emptyset$ is in $\tau$ by definition (trivially). Since $\mathbb{R}^2$ is unbounded, then for every unit vector $\textbf{u}$ there always exists $\epsilon>0$ such that $\textbf{x}+s\textbf{u}\in \mathbb{R}^2$. So $\mathbb{R}^2 \in \tau$.
- Let $\{U_\alpha\}_{\alpha \in J} \in \tau$, and let $\textbf{x} \in \bigcup_{\alpha \in J}U_\alpha$. Then $\textbf{x} \in U_\alpha$ for some $\alpha \in J$. Since $U_\alpha \in \tau$, for every unit vector $\textbf{u}$, there exists $\epsilon>0$ such that $\textbf{x}+s\textbf{u}\in U_\alpha$. Thus for every unit vector $\textbf{u}$, there exists $\epsilon>0$ such that $\textbf{x}+s\textbf{u}\in \bigcup_{\alpha \in J} U_\alpha$. Hence $\bigcup_{\alpha \in J} U_\alpha \in \tau$.
- Let $\textbf{x} \in \bigcap_{i=1}^n U_i$. Then $\textbf{x} \in U_i$ for each $i = 1,...,n$. Since each $U_i \in \tau$, for every unit vector $\textbf{u}$, there exists $\epsilon_i>0$ such that $\textbf{x}+s\textbf{u}\in U_i$. Take $\epsilon = \min\{\epsilon_i\}$. It follows that for every unit vector $\textbf{u}$, there exists $\epsilon>0$ such that $\textbf{x}+s\textbf{u}\in \bigcap_{i=1}^n U_i$. So $\bigcap_{i=1}^n U_i \in \tau$.